Recall the distance formula:
\[d^2=(x-x_0)^2+(y-y_0)^2\]
In this particular problem, you have \((x_0,y_0)=(0,0)\) (the origin), since you want to find the distance between any given point on the curve \((x,y)\) and the origin.
\[d^2=x^2+y^2\]
Substitute the given equation of the curve:
\[d^2=x^2+\left(4\sqrt{2x+2}\right)^2\\
d^2=x^2+16(2x+2)\\
d^2=x^2+32x+32\]
Now differentiate both sides implicitly with respect to some dummy variable \(t\):
\[2d\frac{dd}{dt}=(2x+32)\frac{dx}{dt}\]
You're told that at the point \((1,8)\), \(x\) is increasing at a rate of 3 units/s, which translates to \(\dfrac{dx}{dt}=3\).
\[2d\frac{dd}{dt}=(2\cdot1+32)(3)\\
2d\frac{dd}{dt}=102\]
You're asked to find \(\dfrac{dd}{dt}\), so first you need to find \(d\). To do this, find the distance between \((1,8)\) and \((0,0)\), then plug it into the above equation. Lastly, solve for \(\dfrac{dd}{dt}\).